FIGURE CP33.73 shows two nearly overlapped intensity peaks | StudySoup

Textbook Solutions for Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36)

Chapter 33 Problem 33.95

Question

FIGURE CP33.73 shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating (see Figure 33.9b). As a practical matter, two peaks can just barely be resolved if their spacing y equals the width w of each peak, where w is measured at half of the peaks height. Two peaks closer together than w will merge into a single peak. We can use this idea to understand the resolution of a diffraction grating. a. In the small-angle approximation, the position of the m = 1 peak of a diffraction grating falls at the same location as the m = 1 fringe of a double slit: y1 = lL/d. Suppose two wavelengths differing by l pass through a grating at the same time. Find an expression for y, the separation of their first-order peaks. b. We noted that the widths of the bright fringes are proportional to 1/N, where N is the number of slits in the grating. Lets hypothesize that the fringe width is w = y1/N. Show that this is true for the double-slit pattern. Well then assume it to be true as N increases. c. Use your results from parts a and b together with the idea that ymin = w to find an expression for lmin, the minimum wavelength separation (in first order) for which the diffraction fringes can barely be resolved. d. Ordinary hydrogen atoms emit red light with a wavelength of 656.45 nm. In deuterium, which is a heavy isotope of hydrogen, the wavelength is 656.27 nm. What is the minimum number of slits in a diffraction grating that can barely resolve these two wavelengths in the first-order diffraction pattern?

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The first step in solving 33 problem number 95 trying to solve the problem we have to refer to the textbook question: FIGURE CP33.73 shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating (see Figure 33.9b). As a practical matter, two peaks can just barely be resolved if their spacing y equals the width w of each peak, where w is measured at half of the peaks height. Two peaks closer together than w will merge into a single peak. We can use this idea to understand the resolution of a diffraction grating. a. In the small-angle approximation, the position of the m = 1 peak of a diffraction grating falls at the same location as the m = 1 fringe of a double slit: y1 = lL/d. Suppose two wavelengths differing by l pass through a grating at the same time. Find an expression for y, the separation of their first-order peaks. b. We noted that the widths of the bright fringes are proportional to 1/N, where N is the number of slits in the grating. Lets hypothesize that the fringe width is w = y1/N. Show that this is true for the double-slit pattern. Well then assume it to be true as N increases. c. Use your results from parts a and b together with the idea that ymin = w to find an expression for lmin, the minimum wavelength separation (in first order) for which the diffraction fringes can barely be resolved. d. Ordinary hydrogen atoms emit red light with a wavelength of 656.45 nm. In deuterium, which is a heavy isotope of hydrogen, the wavelength is 656.27 nm. What is the minimum number of slits in a diffraction grating that can barely resolve these two wavelengths in the first-order diffraction pattern?
From the textbook chapter Wave Optics you will find a few key concepts needed to solve this.

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Title Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36) 4 
Author Randall D. Knight (Professor Emeritus)
ISBN 9780134081496

FIGURE CP33.73 shows two nearly overlapped intensity peaks

Chapter 33 textbook questions

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